Find the vertices, foci, and eccentricity of the ellipse. 5x^2+y^2=6 can someone help me with this?

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Find the vertices, foci, and eccentricity of the ellipse. 5x^2+y^2=6 can someone help me with this?

Mathematics
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when x = 0, y = +-sqrt(6) when y = 0, x = +-sqrt(6/5) the rest of it is greek to me at the moment :)
it appears to be taller than wider
those might be the vertices :)

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i get how to do the problem. It should look like x^2/6/5 + y^2/30 = 1. I just dont know how to clear the fraction under the x^2 term.
1/(6/5) just means find the reciprocal of 6/5
the equation must stay equal to 1.
foci are c^2 = a^2 - b^2....
[(x-h)^2]/a+[(y-k)^2}/b=1 what is this form?
5^2 - 1^2 = 24.... is the slant = sqrt(24)? cant recall to much about it
i remember writing ellipses in that form
do the foci need to be on the x axis?
yes because it is a horizontal equation
heres some information on horizontal ellipses is x^2/a^2 + y^2/b^2 = 1 a> b > 0 vertices (+ or -a,0) foci (+ or - c,0)
I know everything just not how to clear that 6/5 without changing the = 1 part.
if i could just put 5x^2+y^2=6 in the x^2/a^2 + y^2/b^2 = 1 form. i could solve it.
I would just leave the 6/5 where it is. But I think there is a 6 under the y^2, not 30.
your right
What do you mean "solve"? You say you know how to do it...
I mean find all the information from it and then graph it.
Okay, the vertices are (+-sqrt(6/5), 0) and (0, +-sqrt(6)). The foci you find by taking c^2 = a^2 - b^2. So, c^2 = 6 - 6/5 = 24/5. c = +-sqrt(24/5). The foci are on the y-axis because the ellipse is vertically oriented: (0, +-sqrt(24/5)). The eccentricity you find by taking c / a, so sqrt(24/5) / sqrt(6) = sqrt(4/5).
Cool. thats what i got. Thanx!

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